Static model of a 2×25kV AC traction system

Plakhova, Mariia; Belaid, Mohamed; Arboleya, Pablo

doi:10.1109/pesa.2015.7398916

Cited by 5 publications

(5 citation statements)

References 4 publications

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“…Tus, the equivalent catenary and the equivalent feeder of Figure 1 do not represent in detail the catenaries and feeders of each railway in the same path. Despite this fact, simplifed schemes (such as the one shown in Figure 1) have been frequently analyzed in the literature about this point [3,4,[6][7][8][9][10][13][14][15][16].…”

Section: Brief Description Of 2 × 25 Kv Electric Railway Systemsmentioning

confidence: 99%

“…At each point of the railway path, the current in the catenary must be the sum of the current in the feeder plus the current through the ground paths (rails, ground conductors, and physical earth). Tat is, for cells 1, 2, and 3, the current through the ground path is not null (and this point has been sometimes oversimplifed [3][4][5][6][7][8][9]). Te equation ( 5) clearly shows that I F is equal to I C only if Z C is equal to Z F (i.e., for type-3 cells, the current by the ground path can only be null if Z C is equal to Z F ).…”

Section: Distribution Of Currents For Cases With Only One Trainmentioning

confidence: 99%

“…Te distributions of currents in 2 × 25 kV electric railway systems have been sometimes described in an oversimplifed way [4][5][6][7][8][9], but there are also documents that consider that those distributions of currents are not so simple [10][11][12][13][14][15][16][17][18][19][20][21][22]. Some simplifed distributions, dependent on train position, are described in [10,11] without considering the mutualimpedances between the catenary and feeder.…”

Section: Introductionmentioning

confidence: 99%

“…Tus, the simple question about how the current of one train is distributed through the main paths (equivalent catenary, equivalent feeder, and ground) remains without a simple and proper answer after the review of the literature on this topic. Only some simplistic answers [4][5][6][7][8][9] were available (without solving the load fow problem), but such simplistic answers are extremely imprecise (e.g., in some references, the train current is multiplied by assumed distribution factors, such as 1/4, 3/4, and/or 1/2, regardless of train position). Tis article ofers a straightforward answer to that question, based on simple analytic equations that are also useful for an approximate solution to the load fow problem.…”

Section: Introductionmentioning

confidence: 99%

See 3 more Smart Citations

Distribution of Currents in 2 × 25 kV Electric Railway Systems under Normal Conditions

Sorrentino

Gupta

Montilla-DJesus

et al. 2023

International Transactions on Electrical Energy Systems

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This article shows a detailed analysis of the distribution of currents in 2 × 25 kV electric railway systems under normal conditions. The network equations are clearly formulated in a way that enables the presentation of a novel simplified analytical solution as well as the traditional numerical solution of the equation set. The simplified analytical solution is obtained when the transformer impedances are neglected, and under these conditions: (a) the distributions of currents are analytically deduced for cases with only one train; (b) the distribution of currents among autotransformers and between catenary and feeder can be easily understood, as well as the effect of the train position on the distribution of currents; and (c) the superposition method is applied for cases with multiple trains in order to clearly explain the distribution of currents from the results with only one train. On the other hand, the network equations are also numerically solved, including autotransformer impedances, and it is shown that their effect is very low, especially because these impedances are typically small. Therefore, the proposed analytical method is a good tool to obtain an easy and approximate solution for the distribution of currents in these systems, as well as an excellent tool to facilitate the understanding of that distribution.

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Section: Brief Description Of 2 × 25 Kv Electric Railway Systemsmentioning

confidence: 99%

Section: Distribution Of Currents For Cases With Only One Trainmentioning

confidence: 99%

Section: Introductionmentioning

confidence: 99%

Section: Introductionmentioning

confidence: 99%

See 2 more Smart Citations

Distribution of Currents in 2 × 25 kV Electric Railway Systems under Normal Conditions

Sorrentino

Gupta

Montilla-DJesus

et al. 2023

International Transactions on Electrical Energy Systems

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show abstract

“…Compared to the circulating in the secondary windings of the transformer. Compared to the classical twowire 25 kV railway electrification system, the voltage drops and the Joule losses are lowered [3,4]. This is why the "2 × 25 kV" system is commonly used worldwide for supplying railway high-speed lines.…”

Section: Introductionmentioning

confidence: 99%

Calculation of the Voltage Unbalance Factor for High-Speed Railway Substations with V-Connection Scheme

2022

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In France, high-speed railway lines are powered by a 2 × 25 kV/50 Hz electrification system. The substations include two single-phase transformers connected to the high-voltage electrical transmission network on different pairs of phases according to a so-called “V-connection scheme”. In practice, due to the large variations in the power absorbed by the trains, this connection does not make it possible to satisfactorily limit the unbalance in the three-phase voltages. In order to correctly size a balancing system to be associated with the substation, it is necessary to calculate, with precision, the voltage unbalance factor as a function of the power drawn by the trains. In its first part, this paper presents modelling of the substation and proposes an algorithm which allows for the calculation of the upstream line voltage as a function of the power consumption at the secondary of the transformers. The voltage unbalance factor can then be determined over a long period of operation. In the second part of this paper, the same approach is applied with an unbalance-compensator based on Steinmetz circuits controlled by AC choppers. Finally, in both cases, the results of the calculations are validated by simulations performed with PLECS simulation software.

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